What volume of sulfur dioxide is formed during the combustion of 67.2 liters of hydrogen sulphide?

Given:

V (H2S) = 67.2 l

V (SO2) -?

Solution:

Let’s find the amount of substance H2S.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

n = V: Vn.

n = 67.2 L: 22.4 L / mol = 3 mol.

Let’s find the quantitative ratios of substances.

2H2S + 3О2 = 2SО2 + 2Н2О

For 2 mol of H2S, there are 2 mol of SO2.

Substances are in quantitative ratios 1: 1.

The amount of substances will be equal.

n (SO2) = n (H2S) = 3 mol.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 3 mol × 22.4 L / mol = 67.2 L.

Answer: 67.2 liters.